Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
double sy, a, b;
struct gvector {
  double x, y;
  gvector(){};
  gvector(double x, double y) {
    this->x = x;
    this->y = y;
  }
  gvector operator+(gvector t) {
    gvector n = gvector(*this);
    n.x += t.x;
    n.y += t.y;
    return n;
  }
  gvector operator-(gvector t) {
    gvector n = gvector(*this);
    n.x -= t.x;
    n.y -= t.y;
    return n;
  }
  gvector operator*(double k) {
    gvector n = gvector(*this);
    n.x *= k;
    n.y *= k;
    return n;
  }
  double length() { return sqrt(x * x + y * y); }
  void show() { cout << "  x=" << x << " y=" << y << endl; }
};
double fencel[200005];
double fencer[200005];
double pre[200005];
int main() {
  while (cin >> sy >> a >> b) {
    int n;
    double l, r;
    gvector A(a, sy);
    gvector B(b, sy);
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
      scanf("%lf%lf", &fencel[i], &fencer[i]);
      pre[i + 1] = pre[i] + (fencer[i] - fencel[i]);
    }
    int q;
    double x, y;
    scanf("%d", &q);
    for (int i = 0; i < q; i++) {
      scanf("%lf%lf", &x, &y);
      gvector p(x, y);
      double sfb = (y - 0) / (y - sy);
      gvector pa = A - p;
      gvector pb = B - p;
      pa = pa * sfb;
      pb = pb * sfb;
      pa = p + pa;
      pb = p + pb;
      if (pb.x <= fencel[0]) {
        printf("0.000000000000000\n");
        continue;
      }
      if (pa.x >= fencer[n - 1]) {
        printf("0.000000000000000\n");
        continue;
      }
      int l = 0, r = n - 1, mid;
      while (l <= r) {
        mid = (l + r) / 2;
        if (fencer[mid] > pa.x + 1e-8)
          r = mid - 1;
        else
          l = mid + 1;
      }
      mid = (l + r) / 2;
      l = r + 1;
      double left = 0;
      left = pre[l + 1];
      left -= min(fencer[l] - fencel[l], (fencer[l] - pa.x));
      l = 0, r = n - 1;
      while (l <= r) {
        mid = (l + r) / 2;
        if (fencel[mid] < pb.x - 1e-8)
          l = mid + 1;
        else
          r = mid - 1;
      }
      mid = (l + r) / 2;
      r = l - 1;
      double right = 0;
      right = pre[n] - pre[r];
      right -= min(fencer[r] - fencel[r], pb.x - fencel[r]);
      double inter = pre[n] - right - left;
      double pro = inter / (pa - pb).length();
      double time = (b - a) * pro;
      printf("%.15f\n", time);
    }
  }
  return 0;
}